Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $p \neq 0$. $k = \dfrac{2p + 6}{-9p - 54} \times \dfrac{-7p^2 - 42p}{p^2 + 2p - 3} $
First factor the quadratic. $k = \dfrac{2p + 6}{-9p - 54} \times \dfrac{-7p^2 - 42p}{(p + 3)(p - 1)} $ Then factor out any other terms. $k = \dfrac{2(p + 3)}{-9(p + 6)} \times \dfrac{-7p(p + 6)}{(p + 3)(p - 1)} $ Then multiply the two numerators and multiply the two denominators. $k = \dfrac{ 2(p + 3) \times -7p(p + 6) } { -9(p + 6) \times (p + 3)(p - 1) } $ $k = \dfrac{ -14p(p + 3)(p + 6)}{ -9(p + 6)(p + 3)(p - 1)} $ Notice that $(p + 6)$ and $(p + 3)$ appear in both the numerator and denominator so we can cancel them. $k = \dfrac{ -14p\cancel{(p + 3)}(p + 6)}{ -9(p + 6)\cancel{(p + 3)}(p - 1)} $ We are dividing by $p + 3$ , so $p + 3 \neq 0$ Therefore, $p \neq -3$ $k = \dfrac{ -14p\cancel{(p + 3)}\cancel{(p + 6)}}{ -9\cancel{(p + 6)}\cancel{(p + 3)}(p - 1)} $ We are dividing by $p + 6$ , so $p + 6 \neq 0$ Therefore, $p \neq -6$ $k = \dfrac{-14p}{-9(p - 1)} $ $k = \dfrac{14p}{9(p - 1)} ; \space p \neq -3 ; \space p \neq -6 $